Explanation:

A transition metal complex absorbs visible tight only if it has unpaired electrons.
No unpaired electron so does not absorb visible light.

Explanation:

Key ldea In case of high spin complex, \(\Delta_{0}\) is small. Thus, the energy required to pair up the fourth electron with the
electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital.
Thus, pairing does not occur.
For high spin \(d^{4}\) octahedral complex Crystal fleld stabilisation energy
\(=(-3 \times 0.4+1 \times 0.6) \Delta_{0}\)
\(=(-1.2+0.6) \Delta_{0}=-0.6 \Delta_{0}\)

Explanation:

Complexes of \(\left[M A_{4} B_{2}\right]\) type exhibit geometrical isomerism. The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]+is\; \mathrm{a}\;\left[M A_{4} B_{2}\right]\) type complex and thus, fulfills the conditions that are necessary to exhibit geometrical isomerism. Hence, it has two geometrical isomers of different colours as: The structure of the geometrical isomers is as For linkage isomerism, presence of ambidentate ligand is necessary. For coordination isomerism, both the cation and anion of the complex must be complex ions. For lonisation isomerism, an nion different to the ligands must be present outside the coordination sphere. All these conditions are not satisfled by this complex. Hence, it does not exhibit other given isomerisms.

Explanation:

Explanation:

A transition metal complex absorbs visible tight only if it has unpaired electrons.
No unpaired electron so does not absorb visible light.

Explanation:

Key ldea In case of high spin complex, \(\Delta_{0}\) is small. Thus, the energy required to pair up the fourth electron with the
electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital.
Thus, pairing does not occur.
For high spin \(d^{4}\) octahedral complex Crystal fleld stabilisation energy
\(=(-3 \times 0.4+1 \times 0.6) \Delta_{0}\)
\(=(-1.2+0.6) \Delta_{0}=-0.6 \Delta_{0}\)

Explanation:

Complexes of \(\left[M A_{4} B_{2}\right]\) type exhibit geometrical isomerism. The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]+is\; \mathrm{a}\;\left[M A_{4} B_{2}\right]\) type complex and thus, fulfills the conditions that are necessary to exhibit geometrical isomerism. Hence, it has two geometrical isomers of different colours as: The structure of the geometrical isomers is as For linkage isomerism, presence of ambidentate ligand is necessary. For coordination isomerism, both the cation and anion of the complex must be complex ions. For lonisation isomerism, an nion different to the ligands must be present outside the coordination sphere. All these conditions are not satisfled by this complex. Hence, it does not exhibit other given isomerisms.

Explanation:

Explanation:

A transition metal complex absorbs visible tight only if it has unpaired electrons.
No unpaired electron so does not absorb visible light.

Explanation:

Key ldea In case of high spin complex, \(\Delta_{0}\) is small. Thus, the energy required to pair up the fourth electron with the
electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital.
Thus, pairing does not occur.
For high spin \(d^{4}\) octahedral complex Crystal fleld stabilisation energy
\(=(-3 \times 0.4+1 \times 0.6) \Delta_{0}\)
\(=(-1.2+0.6) \Delta_{0}=-0.6 \Delta_{0}\)

Explanation:

Complexes of \(\left[M A_{4} B_{2}\right]\) type exhibit geometrical isomerism. The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]+is\; \mathrm{a}\;\left[M A_{4} B_{2}\right]\) type complex and thus, fulfills the conditions that are necessary to exhibit geometrical isomerism. Hence, it has two geometrical isomers of different colours as: The structure of the geometrical isomers is as For linkage isomerism, presence of ambidentate ligand is necessary. For coordination isomerism, both the cation and anion of the complex must be complex ions. For lonisation isomerism, an nion different to the ligands must be present outside the coordination sphere. All these conditions are not satisfled by this complex. Hence, it does not exhibit other given isomerisms.

Explanation:

Explanation:

A transition metal complex absorbs visible tight only if it has unpaired electrons.
No unpaired electron so does not absorb visible light.

Explanation:

Key ldea In case of high spin complex, \(\Delta_{0}\) is small. Thus, the energy required to pair up the fourth electron with the
electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital.
Thus, pairing does not occur.
For high spin \(d^{4}\) octahedral complex Crystal fleld stabilisation energy
\(=(-3 \times 0.4+1 \times 0.6) \Delta_{0}\)
\(=(-1.2+0.6) \Delta_{0}=-0.6 \Delta_{0}\)

Explanation:

Complexes of \(\left[M A_{4} B_{2}\right]\) type exhibit geometrical isomerism. The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]+is\; \mathrm{a}\;\left[M A_{4} B_{2}\right]\) type complex and thus, fulfills the conditions that are necessary to exhibit geometrical isomerism. Hence, it has two geometrical isomers of different colours as: The structure of the geometrical isomers is as For linkage isomerism, presence of ambidentate ligand is necessary. For coordination isomerism, both the cation and anion of the complex must be complex ions. For lonisation isomerism, an nion different to the ligands must be present outside the coordination sphere. All these conditions are not satisfled by this complex. Hence, it does not exhibit other given isomerisms.

Explanation: